Instantaneous Power — Revision Notes

⚡ 30-Second Revision

Definition: — Rate of doing work at an instant.
Formula 1: — $P = \frac{dW}{dt}$
Formula 2 (Key): — $P = \vec{F} \cdot \vec{v}$
Scalar Quantity: — Power has magnitude only.
Units: — Watt (W) or J/s.
Dot Product: — $P = Fv\cos\theta$, where $\theta$ is angle between $\vec{F}$ and $\vec{v}$.
Zero Power: — If $\vec{F} \perp \vec{v}$ (e.g., centripetal force) or $\vec{v} = \vec{0}$.
Negative Power: — If force component is opposite to velocity ($\theta > 90^\circ$).
Relation to Kinetic Energy: — $P_{net} = \frac{dK}{dt}$.

2-Minute Revision

Instantaneous power is the rate at which work is done or energy is transferred at a precise moment in time. It is fundamentally defined as the time derivative of work done, $P = \frac{dW}{dt}$. The most practical and frequently used formula for instantaneous power is the dot product of the instantaneous force $\vec{F}$ acting on an object and its instantaneous velocity $\vec{v}$, expressed as $P = \vec{F} \cdot \vec{v}$.

This formula highlights that only the component of the force parallel to the velocity contributes to the power. Instantaneous power is a scalar quantity, meaning it has magnitude but no direction, and its SI unit is the Watt (W), equivalent to Joules per second (J/s).

It can be positive (energy added), negative (energy removed), or zero (force perpendicular to velocity or zero velocity). For NEET, remember to use calculus (differentiation) when force or velocity are given as functions of time or position, and be adept at vector dot product calculations.

5-Minute Revision

Instantaneous power is a critical concept in physics, representing the rate of work done or energy transfer at any given instant. It's distinct from average power, which is measured over a time interval.

The core mathematical definition is $P = \frac{dW}{dt}$, where $dW$ is the infinitesimal work done over an infinitesimal time $dt$. A more practical and frequently tested form is $P = \vec{F} \cdot \vec{v}$, the dot product of the instantaneous force vector and the instantaneous velocity vector.

This formula is crucial because it immediately tells us that power is a scalar quantity, and only the component of the force parallel to the velocity contributes to the power. If the angle between $\vec{F}$ and $\vec{v}$ is $\theta$, then $P = Fv\cos\theta$.

Key Points to Remember:

1
Scalar Nature: — Power is always a scalar, even though force and velocity are vectors.
2
Units: — The SI unit is Watt (W), which is $1\text{ J/s}$.
3
Zero Power Conditions: — Power is zero if:

* The object is momentarily at rest ($\vec{v} = \vec{0}$). Example: A ball at the peak of its trajectory. * The force is perpendicular to the velocity ($\vec{F} \perp \vec{v}$, so $\cos\theta = 0$). Example: Centripetal force in uniform circular motion.

1
Negative Power: — If the force component is opposite to the direction of velocity ($\theta > 90^\circ$), power is negative, meaning energy is being removed from the system (e.g., friction, braking force).
2
Calculus Application: — For problems where force or velocity are functions of time or position, you'll need to use differentiation to find instantaneous values. For example, if $x(t)$ is given, find $v(t) = \frac{dx}{dt}$ and $a(t) = \frac{dv}{dt}$, then $F(t) = ma(t)$, and finally $P(t) = F(t)v(t)$.

Worked Example: A $1\text{ kg}$ object moves such that its velocity is given by $\vec{v} = (2t\hat{i} + 3\hat{j})\text{ m/s}$. A constant force $\vec{F} = (4\hat{i} - 2\hat{j})\text{ N}$ acts on it. Find the instantaneous power at $t = 1\text{ s}$.

Solution:

1
Find velocity at $t = 1\text{ s}$: $\vec{v}(1) = (2(1)\hat{i} + 3\hat{j}) = (2\hat{i} + 3\hat{j})\text{ m/s}$.
2
The force is constant: $\vec{F} = (4\hat{i} - 2\hat{j})\text{ N}$.
3
Calculate instantaneous power: $P = \vec{F} \cdot \vec{v}(1) = (4)(2) + (-2)(3) = 8 - 6 = 2\text{ W}$.

This example demonstrates the direct application of the dot product formula, which is a common NEET question type.

Prelims Revision Notes

Instantaneous power ($P$) is the rate of work done or energy transfer at a specific moment. It is defined as $P = \frac{dW}{dt}$. The most frequently used formula is $P = \vec{F} \cdot \vec{v}$, where $\vec{F}$ is the instantaneous force and $\vec{v}$ is the instantaneous velocity.

Remember that power is a scalar quantity, always. Its SI unit is the Watt (W), which is equivalent to Joules per second (J/s). When force and velocity are given as vectors, calculate the dot product by multiplying corresponding components and summing them: $P = F_x v_x + F_y v_y + F_z v_z$.

If the magnitudes and the angle $\theta$ between them are known, use $P = Fv\cos\theta$.

Crucial conditions for power:

Zero Power: — Occurs when $\vec{F}$ is perpendicular to $\vec{v}$ (e.g., centripetal force, normal force on a horizontal surface) or when $\vec{v} = \vec{0}$ (e.g., at the highest point of projectile motion).
Negative Power: — Occurs when the component of force is opposite to the direction of velocity ($\theta > 90^\circ$), indicating energy is being removed from the system.

For problems involving time-dependent position or velocity:

1
If $x(t)$ is given, find $v(t) = \frac{dx}{dt}$.
2
If $v(t)$ is given, find $a(t) = \frac{dv}{dt}$.
3
Use Newton's second law, $F = ma$, to find the force if mass is given.
4
Then apply $P = Fv$ (if parallel) or $P = \vec{F} \cdot \vec{v}$ (for vectors) at the specified time instant.

Also, recall the connection to the work-energy theorem: the instantaneous power delivered by the net force is equal to the rate of change of kinetic energy, $P_{net} = \frac{dK}{dt}$. This means if $K(t)$ is given, you can find $P(t)$ by differentiating $K(t)$ with respect to time. Practice problems involving all these scenarios to ensure quick recall and accurate application.

Vyyuha Quick Recall

Power Is Force Velocity Dot-product. (P = F \cdot v)